Question 2010 pts20. Calculate the change in internal energy (AE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.00 L in volumeat 1.50 atm pressure.(Remember that 101.3J = 1 Latm)+25.9 kJ○ -16.0 kJ-25.9 kJO-24.1 kJ+937 kJ

To solve for the change in internal energy (ΔE) for the given system, we need to use the first law…

Answered: Question 20 10 pts 20. Calculate the…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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ALab Data Temperature of cold water (°C) Temperature of hot water (°C) Volume of cold water (mL) Volume of hot water (mL) Final temperature after mixing (°C) Mass of cold water (g) Mass of hot water (g) Calorimeter constant (J/°C) 4.0 87.0 98.0 83.0 41.0 98.0 83.0 How to calculate the calorimeter constant Sm -
200 mL of a 0.500 M solution of AgNO, reacts with 200 mL of a 0.500 M solution of HBr in a coffee-cup calorimeter and the temperature rises from 22.80 °C to 26.58 °C. Assuming the density and specific heat of the resulting solution is 1.00 g/mL and 4.184 J/g.°C, respectfully, what is the AH°rxn? O.45.6kJ/mol O+57.3 kJ/mol O- 63.3 kJ/mol
Calculate the heat value for a 1.3 g Tungsten wire which is heated from 25.0°C to 50.3°C? Consider the specific heat capacity of Tungsten is 0.152 J g-10C¯1. 1.7 0.004
4. The reaction of 250.0 mL of a 1.00 M hydrochloric acid solution with 250.0 mL of a 1.00 M sodium hydroxide solution was carried out in a constant pressure calorimeter. The total heat capacity of the calorimeter plus solutions was 6.45 kJ/K. The temperature of the calorimeter and solutions increased by 2.11°C. What is AH (in kJ) for the neutralization of 1.00 mol HCl(aq) by NaOH(aq)? A) -54.4 B) -21.2 +12.6 +54.4 E) -12.6
Find the total energy involved in converting 450.0g of water from -21°C to 35°C. Hvan Water = 40.7 KJ/mol Hrus Water = 6.01 KJ/mol Solid C =2.03 J/g°C water Liquid Cwater =4.184 J/g°C Gas Cwater 2.01 J/g°C O 235 KJ O 150 KJ O 85 KJ 255 KJ
One day while camping, a student used a propane stove to boil some water. The student heated 5.00 L of water from 15.6°C to 98.5°C. The molar enthalpy of combustion of propane is -2043.9 kJ/mol and heating the water required 1.38 mol of propane. Determine the efficiency of the propane camp stove if the heat released from the stove was used to heat the water. 62.4 % 38.4% D162 % 61.6%
Please answer question 20 part A
SUBJECT : GENERAL CHEMISTRY 2 TOPIC : HEAT AND CALORIMETRY
To calibrate a calorimeter, 10.0 ml of 1.0 M NaOH and 5.0 ml of 1.0 M HCl was used in a calorimetry experiment. Which of the following will have NO EFFECT on the calculated heat capacity of the calorimeter, Ccal? a. Accidentally adding HNO3 instead of HCl. b.Adding a smaller volume of 1.0 M HCl. c.Using a bigger styrofoam ball for calibration. d.Using a wet test tube for calibration.
pt pt pt pt pt pt pt pt pt pt pt pt pt pt You should use care when dissolving H₂SO4 in water because the process is highly exothermic. To measure the enthalpy change, 5.8 g of concentrated H₂SO4() was added (with stirring) to 140. g of water in a coffee-cup calorimeter. This resulted in an increase in temperature from 21.0 °C to 30.3 °C. Calculate the enthalpy change for the process H₂SO4(l) → H₂SO4 (aq), in kJ/mol. Assume the specific heat capacity of the solution is 4.2 J/g. K. Enthalpy change = kJ/mol ?
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Question 17 of 35 Submit When 1.151 grams of sucrose (Molar mass 342.3 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter increases from 22.41°C to 26.63°C. If the heat capacity of the calorimeter is 4.900 kJ/°C, what is the heat of combustion of sucrose? kJ/mol 1 3 4 C 7 +/- x 10 0 Tap here or pull up for additional resources LO 00

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Question 20 10 pts 20. Calculate the change in internal energy (AE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.00 L in volume at 1.50 atm pressure. (Remember that 101.3J = 1 Latm) +25.9 kJ ○ -16.0 kJ -25.9 kJ O-24.1 kJ +937 kJ